Math Problem Statement

g(t)=1170*(1.1197)^t. f(4)f(0)f(4)f(0)

f(2)f(1)f(2)f(1)

f(1)f(0)f(1)f(0)

f(2)−f(1)2−1f(2)-f(1)2-1

f(4)−f(0)2−0f(4)-f(0)2-0 f(3)−f(1)3−1

Solution

Let's break down what each of these expressions represents and how they relate to the exponential model.

1. f(4)f(0)\frac{f(4)}{f(0)}

This expression represents the ratio of the number of new cases on day 4 (June 5) to the number of new cases on day 0 (June 1). For an exponential model, this would give us the overall growth factor raised to the power of 4:

f(4)f(0)=(1.1197)4\frac{f(4)}{f(0)} = (1.1197)^4

2. f(2)f(1)\frac{f(2)}{f(1)}

This represents the ratio of the number of new cases on day 2 (June 3) to the number of new cases on day 1 (June 2). This should give us the growth factor itself since it's the increase from one day to the next:

f(2)f(1)=(1.1197)\frac{f(2)}{f(1)} = (1.1197)

3. f(1)f(0)\frac{f(1)}{f(0)}

This represents the ratio of the number of new cases on day 1 (June 2) to the number of new cases on day 0 (June 1). Like the previous expression, it should also give us the growth factor:

f(1)f(0)=(1.1197)\frac{f(1)}{f(0)} = (1.1197)

4. f(2)f(1)21\frac{f(2) - f(1)}{2 - 1}

This expression is trying to calculate the average rate of change (slope) between day 2 and day 1. However, for exponential growth, the rate of change is not constant, and this expression does not represent the growth factor but the difference between the number of cases on day 2 and day 1:

f(2)f(1)21=g(2)g(1)1\frac{f(2) - f(1)}{2 - 1} = \frac{g(2) - g(1)}{1}

5. f(4)f(0)40\frac{f(4) - f(0)}{4 - 0}

This expression calculates the average rate of change between day 4 (June 5) and day 0 (June 1). Like the previous one, it does not directly represent the growth factor for exponential growth, but rather the difference in the number of cases over a period divided by the number of days:

f(4)f(0)40=g(4)g(0)4\frac{f(4) - f(0)}{4 - 0} = \frac{g(4) - g(0)}{4}

6. f(3)f(1)31\frac{f(3) - f(1)}{3 - 1}

This expression calculates the average rate of change between day 3 (June 4) and day 1 (June 2). Like the other difference expressions, it does not directly represent the growth factor but rather the difference in cases divided by the number of days:

f(3)f(1)31=g(3)g(1)2\frac{f(3) - f(1)}{3 - 1} = \frac{g(3) - g(1)}{2}

Conclusion

The expressions that directly represent the growth factor between consecutive days in an exponential model are:

  • f(2)f(1)\frac{f(2)}{f(1)}
  • f(1)f(0)\frac{f(1)}{f(0)}

These represent the growth factor 1.11971.1197 that we identified earlier.

Follow-Up Questions:

  1. How would you interpret f(4)f(0)\frac{f(4)}{f(0)} in the context of exponential growth?
  2. How does the growth factor change when calculated for different time intervals?
  3. What does the average rate of change between two days tell us about the spread of the virus?
  4. If f(3)=1636f(3) = 1636, what would f(5)f(5) be using the growth factor?
  5. How would you use the exponential model to predict cases on June 10?
  6. Can the average rate of change be used to accurately describe exponential growth?
  7. Why is f(4)f(0)\frac{f(4)}{f(0)} greater than f(2)f(1)\frac{f(2)}{f(1)}?
  8. What is the significance of calculating f(1)f(0)\frac{f(1)}{f(0)} in the context of daily case reporting?

Tip: In exponential growth models, ratios like f(t+1)f(t)\frac{f(t+1)}{f(t)} are crucial for understanding how quickly the quantity is increasing, especially when growth is not linear.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Rate of Change
Ratio and Proportion

Formulas

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Theorems

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Suitable Grade Level

High School